Questions: Tutorial Exercise Evaluate the integral, where E is enclosed by the paraboloid z=4+x^2+y^2, the cylinder x^2+y^2=5, and the xy-plane. Use cylindrical coordinates. ∫∫∫E e^z dV Step 1 In cylindrical coordinates, the paraboloid z=4+x^2+y^2 has the equation z= and the cylinder x^2+y^2=5 has the equation r=

Solution Steps

To evaluate the integral over the region \( E \) enclosed by the given surfaces, we first convert the equations to cylindrical coordinates. The paraboloid \( z = 4 + x^2 + y^2 \) becomes \( z = 4 + r^2 \) in cylindrical coordinates, where \( r = \sqrt{x^2 + y^2} \). The cylinder \( x^2 + y^2 = 5 \) becomes \( r = \sqrt{5} \). The region \( E \) is bounded by \( 0 \leq r \leq \sqrt{5} \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq z \leq 4 + r^2 \). The integral in cylindrical coordinates is then set up as \(\int_0^{2\pi} \int_0^{\sqrt{5}} \int_0^{4+r^2} e^z \, r \, dz \, dr \, d\theta\).

We need to evaluate the integral

\[ \iiint_{E} e^{z} \, dV \]

over the region \( E \) enclosed by the paraboloid \( z = 4 + x^2 + y^2 \) and the cylinder \( x^2 + y^2 = 5 \). In cylindrical coordinates, the equations become \( z = 4 + r^2 \) and \( r = \sqrt{5} \). The limits for \( r \) are from \( 0 \) to \( \sqrt{5} \), for \( \theta \) from \( 0 \) to \( 2\pi \), and for \( z \) from \( 0 \) to \( 4 + r^2 \).

The integral can be expressed as:

\[ \int_0^{2\pi} \int_0^{\sqrt{5}} \int_0^{4+r^2} e^z \, r \, dz \, dr \, d\theta \]

Evaluating the innermost integral with respect to \( z \):

\[ \int_0^{4+r^2} e^z \, dz = e^{4+r^2} - 1 \]

Thus, the integral becomes:

\[ \int_0^{2\pi} \int_0^{\sqrt{5}} (e^{4+r^2} - 1) r \, dr \, d\theta \]

Now we evaluate the integral with respect to \( r \):

\[ \int_0^{\sqrt{5}} (e^{4+r^2} - 1) r \, dr \]

This can be split into two parts:

1. \( \int_0^{\sqrt{5}} e^{4+r^2} r \, dr \)
2. \( \int_0^{\sqrt{5}} r \, dr \)

The first part evaluates to:

\[ \frac{1}{2} e^{4+r^2} \bigg|_0^{\sqrt{5}} = \frac{1}{2} (e^{9} - e^{4}) \]

The second part evaluates to:

\[ \frac{1}{2} r^2 \bigg|_0^{\sqrt{5}} = \frac{5}{2} \]

Combining these results gives:

\[ \int_0^{\sqrt{5}} (e^{4+r^2} - 1) r \, dr = \frac{1}{2} (e^{9} - e^{4}) - \frac{5}{2} \]

Finally, integrating with respect to \( \theta \):

\[ \int_0^{2\pi} \left( \frac{1}{2} (e^{9} - e^{4}) - \frac{5}{2} \right) d\theta = 2\pi \left( \frac{1}{2} (e^{9} - e^{4}) - \frac{5}{2} \right) \]

Final Answer